2
\$\begingroup\$

Reading through "The Art of Electronics", I came to a statement where the writer states that the time constant (t) of RC circuit must be many times bigger than the Length of pulse you want to couple through the capacitor.

These are the words from the book:

you might use a blocking capacitor to couple pulses, or square waves. In such situations you encounter waveform distortion, in the form of “droop” and overshoot (rather than the simple amplitude attenuation and phase shift you get with sinusoidal waves). Thinking in the time domain, the criterion you use to avoid waveform distortion in a pulse of duration T is that the time constant τ=RC>>T.

Imagine a situation where I need to couple 1us rectangular pulse. Following the book, let's take time constant t=100us.

According to my understanding, time constant is the time required for the circuit to respond to the change. So, whenever the rising edge of the pulse comes in, the circuit is going to take 100us to respond. But the pulse would have already died by the time this circuit settles.

I thought this condition is valid since it wont let any fluctuations happen whenever the pulse is in either high or low state. But in the case of switching from one state to another, I think this condition is invalid.

What is the actual thing going behind the scenes?

\$\endgroup\$
3
  • 1
    \$\begingroup\$ "The Art of Electronics" - One of the best books an engineer can trust :) \$\endgroup\$
    – Mitu Raj
    Nov 8 '20 at 17:13
  • 2
    \$\begingroup\$ \$Xc =1/C2\pi f\$ , shorter pulse has higher frequency components, hence capacitor acts like 'short' for them. ie., it simply passes thru. \$\endgroup\$
    – Mitu Raj
    Nov 8 '20 at 17:20
  • 1
    \$\begingroup\$ Yes, I got it @MituRaj. And this book is one of few books I trust. \$\endgroup\$
    – G-aura-V
    Nov 9 '20 at 1:41
4
\$\begingroup\$

That's right.

If the RC circuit has time constant of 100us, it has no chance to react to the 1us pulse, so it will pass on right through.

If on the other hand you needed to pass a 100us pulse with a RC circuit having also 100us time constant, the 100us pulse would have enough time charge up the RC circuit, so the waveform would have droop.

Same thing can be explained in frequency domain as well. If you have an RC high pass filter with 1 kHz cutoff frequency, it can pass 10kHz sine wave almost at the same amplitude as it comes in, but a 1 kHz sine wave would have attenuated by 3dB, because the signal is slow enough to charge the RC.

\$\endgroup\$
6
  • \$\begingroup\$ Shouldn't the circuit react to allow that change? How can the circuit pass a pulse without changing its state from low to high and again high to low? \$\endgroup\$
    – G-aura-V
    Nov 8 '20 at 16:38
  • \$\begingroup\$ @Giga-Byte Sorry I don't follow what is unclear. A 1us pulse can only charge the 100us RC by a very small amount so it is too short to be affected much by the RC. A 100us pulse can charge a 100us RC by very large amount, so the RC affects the pulse a lot. \$\endgroup\$
    – Justme
    Nov 8 '20 at 16:41
  • \$\begingroup\$ Whenever the rising edge comes in, the capacitor must react to this edge. The capacitor can't just sit there doing nothing, It should account for the leading edge, stabilize and fall back down again. \$\endgroup\$
    – G-aura-V
    Nov 8 '20 at 16:48
  • 1
    \$\begingroup\$ For a large capacitor, it takes a lot of charge to change it's voltage. Since you have a pretty large capacitor, compared to the fast signal that is charging it, the capacitor will not get a lot of charge so it does not develop a significant voltage over it during a short pulse, so both capacitor terminals will have almost same voltage step, and for a large enough capacitor and short enough pulse, the capacitor can be thought as to not charge at all, and both terminals of the capacitor have identical step up and step down. A long pulse will have time to charge the capacitor more. \$\endgroup\$
    – Justme
    Nov 8 '20 at 16:51
  • 1
    \$\begingroup\$ Ahhh! I guess I got it. The capacitor doesn't have large enough time to develop voltage across it. So, it acts just like a piece of wire allowing whatever come in to go out. \$\endgroup\$
    – G-aura-V
    Nov 8 '20 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.